lim∫(arctant)∧2dt √(x∧2 1)
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直接对x求导算不出,所以先对t求导,再对x求导
这是参数方程求导x'=t/(1+t^2)y'=1/(1+t^2)x''=[(1+t^2)-t*2t]/(1+t^2)^2=(1-t^2)/(1+t^2)^2y''=-2t/(1+t^2)^2dy/dx
-(t^2+1)/(4t^3)dy/dt=1/(t*t+1)dx/dt=2t/(t*t+1)dy/dx=1/2td^2y/dx^2=[d(1/2t)/dt]*(t*t+1)/2t=-(t^2+1)/(
由洛必达法则,原式=lim(x趋于无穷)(arctanx)^2/(x/√(x^2+1))=lim√(x^2+1)/x*lim(arctanx)^2=1*(π/2)^2=π^2/4
dy/dx=[1-1/(1+t²)]/[2t/(1+t²)]=t/2d²y/dx²=(1/2)*dt/dx=(1/2)/(dx/dt)=(1/2)/[2t/(1
dx/dt=1-2t/(1+t^2)=(1+t^2-2t)/(1+t^2)=(t-1)^2/(1+t^2)dy/dt=1/(1+t^2)y'=1/(t-1)^2dy'/dt=-2/(t-1)^3y"=
原式=limx→0{2/(1+x^2)-[(1-x)/(1+x)]*[1/(1-x)+(1+x)/(1-x)^2]}/n*x^(n-1)=limx→0[2/(1+x^2)-2/(1-x^2)]/n*x
题目最后一个x是否应该为t?如果是,解答如下lim(x→+∞)∫[0,x](arctant)²dt/√(t²+1)=lim(x→+∞)∫[0,x](arctant)²d(
x't=2t/(1+t^2)y't=1-1/(1+t^2)=t^2/(1+t^2)y'=dy/dx=y't/x't=t/2y"=d(y')/dx=d(y')/dt/(dx/dt)=(1/2)/[2t/
求函数f(x)=(0,x)∫(t+1)arctantdt的极值令df(x)/dx=(x+1)arctanx=0得驻点x₁=-1,x₂=0为书写简便,先求不定积分.∫(t+1)a
那个不是定积分?用洛必达法则lim(x->0){[∫(e²-e^x)dx]/x²}=lim(x->0)[(e²-e^x)/2x]=lim(x->0)[-e^x/2)=-e
先分别求出dx/dt和dy/dt,假设A=dx/dt,B=dy/dt然后用B/A得出dy/dx设C=B/A=dy/dxC中只含有t.因此,d^2y/dx^2=C/dt乘以dx/dt的倒数(dt/dx)
n趋向什么呢?假设是无限吧lim[n→∞](n+1)/(n+2)=lim[n→∞](1+1/n)/(1+2/n)=(1+0)/(1+0)=1lim[n→∞](n²-1)/(2n²+
分别算出dx,dy,然后相除就行详见参考资料
dx/dt=2t/(1+t²)dy/dt=1/(1+t²)dy/dx=1/(2t)d(dx/dt)/dt=(2-4t²)/(1+t²)²d(dy/dt
∵dx=2tdt/√(1+t²)dy=dt/(1+t²)∴dy/dx=[2tdt/√(1+t²)]/[dt/(1+t²)]=2t√(1+t²).
x=tany+ln(cosy^2),dy/dx=(dx/dy)^-1=(tany-1)^-2,y"=d(dy/dx)/dy*dy/dx=-2secy^2/(tany-1)^5